Question

Let $A,B,C$ be finite sets. Suppose that $n(A)=10,n(B)=15,n$ $(C)=20,n(A\cap B)=8$ and $n(B\cap C)=9$. Then the possible value of $n(A\cup B\cup C)$ is

• A. 26
• B. 27
• C. 28
• D. Any of the three values 26, 27, 28 is possible

Answer 1

Answer: (A)

$n(A)=10,n(B)=15,n(20),n(A\cap B)=8,n(B\cap C)=9$
possible value of $n$ (AUBUC)
$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$
$=10+15+20-8-9-n(A\cap C)+n(A\cap B\cap C)$
$=28-n(A\cap C)+n(A\cap B\cap C)$
We see here,
$28-n(A\cap C)+n(A\cap B\cap C)⩾0$
$\Rightarrow n(A\cup B\cup C)⩽28...(1)$
We see, $n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$=10+15-8$
$=17$
$n(B\cup C)=n(B)+n(C)-n(B\cap C)$
$=15+20-9$
$=26$
Obviously, $n(A\cup B\cup C)⩾26$ also,
$n(A\cup B\cup C)⩾17...(2)$
$\therefore$ from (1) & (2)
$26⩽n(A\cup B\cup C)⩽28$